Exercise 16.2
Q1. In the following pairs of triangle (Fig. 12 to 15), the lengths of the sides are indicated along sides. By applying SSS condition, determine which are congruent. State the result in symbolic.
Answer:
1) In \(\Delta\)
AB = DE = 4.5 cm (Side)
BC = EF = 6 cm (Side) and
AC = DF = 4 cm (Side)
Therefore, by SSS criterion of congruence, \(\Delta ABC \cong \Delta DEF\)
2)
In \(\Delta\)
AC = AD (Side)
BC = BD (Side) and
AB = AB (Side)
Therefore, by SSS criterion of congruence, \(\Delta ACB \cong \Delta ADB\)
3) In \(\Delta\)
AB = FE (Side)
AD = FC (Side)
BD = CE (Side)
Therefore, by SSS criterion of congruence, \(\Delta ABD \cong \Delta FEC\)
4) In \(\Delta\)
AB = DC (Side)
AO = OC (Side)
BO = OD (Side)
Therefore, by SSS criterion of congruence, \(\Delta ABO \cong \Delta ODC\)
Q2. In figure, AD = DC and AB = BC
(i) Is \(\Delta ABD \cong \Delta CBD\)
(ii) State the three parts of matching pairs you have used to answer (i).
Answer:
Yes \(\Delta\)
AD = DC
AB = BC and
DB = BD
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Q3. In Figure, AB = DC and BC = AD.
(i) Is \(\Delta ABC \cong \Delta CDA\)
(ii) What congruence condition have you used?
(iii) You have used some fact, not given in the question, what is that?
Answer:
We have AB = DC
BC = AD
and AC = AC
Therefore by SSS \(\Delta ABC \cong \Delta CDA\)
We have used Side Side Side congruence condition with one side common in both the triangles.
Yes, have used the fact that AC = CA.
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Q4. In \(\Delta PQR \cong \Delta EFD\)
(i) Which side of \(\Delta PQR\)
(ii) Which angle of \(\Delta PQR\)
Answer:
\(\Delta PQR \cong \Delta EFD\)
(i) Therefore PR = ED since the corresponding sides of congruent triangles are equal.
(ii) \(\angle\)
Q5. Triangles ABC and PQR are both isosceles with AB = AC and PO = PR respectively. If also, AB = PQ and BC = QR, are the two triangles congruent? Which condition do you use?
It \(\angle\)
Answer:
We have AB = AC in isosceles \(\Delta\)
And PQ = PR in isosceles \(\Delta\)
Also, we are given that AB = PQ and QR = BC.
Therefore, AC = PR (AB = AC, PQ = PR and AB = PQ)
Hence, \(\Delta ABC \cong \Delta PQR\)
Now
\(\angle\)
Therefore, \(\angle\)
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Q6. ABC and DBC are both isosceles triangles on a common base BC such that A and D lie on the same side of BC. Are triangles ADB and ADC congruent? Which condition do you use? If \(\angle\)
Answer,
Yes, triangles ADB and ADC are congruent (By SSS)
AB=AC, DB = DC and AD = DA
\(\angle\)
\(\angle\)
\(\angle\)
\(\angle\)
Since \(\Delta\)
\(\angle\)
2 \(\angle\)
\(\angle\)
Since \(\Delta\)
2 \(\angle\)
\(\angle\)
In \(\Delta\)
\(\angle\)
30° + 20° + \(\angle\)
\(\angle\)
\(\angle\)
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Q7. \(\Delta\)
(i) \(\Delta ABC \cong \Delta ABD\)
(ii) \(\Delta ABC \cong \Delta ADB\)
(iii) \(\Delta ABC \cong \Delta BAD\)
Answer:
In \(\Delta\)
AC = BD (given)
BC = AD (given)
and AB = BA (common)
Therefore by SSS criterion of congruency, Â \(\Delta ABC \cong \Delta BAD\)
There option (iii) is true.
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Q8. In Figure, \(\Delta\)
(i) Is \(\Delta ADB \cong \Delta ADC\)
(ii) State the three pairs of matching parts you use to arrive at your answer.
Answer:
We have AB = AC.
Also since D is the midpoint of BC, BD = DC
Also, AD = DA
Therefore by SSS condition,
\(\Delta ADB \cong \Delta ADC\)
We have used AB, AC : BD, DC AND AD, DA
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Q9. In figure, \(\Delta\)
Answer:
Yes, \(\Delta ABC \cong \Delta ACB\)
Since, ABC is an isosceles triangle, AB = BC, BC = CB and AC = AB
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Q10. Triangles ABC and DBC have side BC common, AB = BD and AC = CD. Are the two triangles congruent? State in symbolic form, which congruence do you use? Does \(\angle\)
Answer:
Yes,
Given,
\(\Delta\)
By SSS criterion of congruency, \(\Delta ABC \cong \Delta DBC\)
No, \(\angle\)